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Math Help - How do I show that this series diverges?

  1. #1
    s3a
    s3a is offline
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    How do I show that this series diverges?

    Again, I am having trouble proving what I keep getting right by using by subconscious logic.

    For the problem:
    a_1 = 2, a_(n+1) = (5n+1)/(4n+3) * a_n ==> Does this converge (absolutely or conditionally) or diverge?

    What I did was:

    the limit of (5n+1)/(4n+3) as n->inf = 5/4 therefore you will be adding increaing numbers and that means the sum will be infinite. Also, to try to prove this, I wrote that when n>=2, 5n+1 >= 4n+3 to show that I am always multiplying the previous number by something equal to or larger than 1 since the sum starts at 2 due to a_1 being predetermined.

    If someone could please help me get my thoughts into mathematical writing then it would be greatly appreciated!
    Thanks in advance!
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  2. #2
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    Quote Originally Posted by s3a View Post
    Again, I am having trouble proving what I keep getting right by using by subconscious logic.

    For the problem:
    a_1 = 2, a_(n+1) = (5n+1)/(4n+3) * a_n ==> Does this converge (absolutely or conditionally) or diverge?

    What I did was:

    the limit of (5n+1)/(4n+3) as n->inf = 5/4 therefore you will be adding increaing numbers and that means the sum will be infinite. Also, to try to prove this, I wrote that when n>=2, 5n+1 >= 4n+3 to show that I am always multiplying the previous number by something equal to or larger than 1 since the sum starts at 2 due to a_1 being predetermined.

    If someone could please help me get my thoughts into mathematical writing then it would be greatly appreciated!
    Thanks in advance!
    You were close...but not quite: as you sequence is positive you can use one of the well-known tests for convergence of positive series, say...D'Alembert's test,

    also known as the ratio test:

    \frac{a_{n+1}}{a_n} = \frac{5n+1}{4n+3}\xrightarrow [n\to\infty]{}\frac{5}{4}>1\Longrightarrow the series diverges.

    Tonio
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